SOLUTION: A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from home to college is 8kn, and she makes the trip in 1 hr. How far does t
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Question 111616: A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from home to college is 8kn, and she makes the trip in 1 hr. How far does the student jog? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r
Let d=distance the student jogs
Then 8-d=distance the student walks
Time spent walking=(8-d)/5
Time spent jogging=d/9
Now we are told that the time spent walking plus the time spent jogging = 1 hour. So:
(8-d)/5+d/9=1 multiply each term by 45 (LCM)
9(8-d)+5d=45 get rid of parens
72-9d+5d=45 subtract 72 from both sides
72-72-9d+5d=45-72 collect like terms
-4d=-27 divide both sides by -4
d=6.75 mi-----------------------distance student jogs
8-6.75=1.25 mi--------------------distance student walks
CK
6.75/9+1.25/5=1
0.75+0.25=1
1=1
